Problem 2 - Master Resource Book in JEE Main Physics

Question

A radio can tune over the frequency range of a portion of MW broadcast bond; \((800 \mathrm{kHz}\) to 1200 \(\mathrm{kHz}\) ). If its LC circuit has an effective inductance of \(220 \mu \mathrm{H}\), what must be the range of its variable capacitor? [Hint For tuning the natural frequency i.e., the frequency of free oscillations of the \(L C\) circuit should be equal to the frequency of the radiowave.] (a) \(87.8\) to \(198 \mathrm{pF}\) (b) 99 to \(190 \mathrm{pF}\) (c) 63 to \(168 \mathrm{pF}\) (d) 44 to \(208 \mathrm{pF}\)

Step-by-Step Solution

Answer

The range of the variable capacitor is 87.8 to 198 pF.

Step 1: Understanding the LC Circuit

An LC circuit, which consists of an inductor (L) and a capacitor (C), resonates at a frequency given by the formula \( f = \frac{1}{2\pi \sqrt{LC}} \). Here, \( f \) is the frequency, \( L \) is the inductance in henrys, and \( C \) is the capacitance in farads.

Step 2: Identify the Frequency Range

The frequency range of the MW broadcast is from \(800 \text{ kHz} \) to \(1200 \text{ kHz} \). Convert these frequencies to hertz: \( f_1 = 800000 \text{ Hz} \) and \( f_2 = 1200000 \text{ Hz} \).

Step 3: Express Capacitance in Terms of Frequency and Inductance

Rearrange the formula \( f = \frac{1}{2\pi \sqrt{LC}} \) to solve for capacitance: \( C = \frac{1}{(2\pi f)^2 L} \). This formula will allow us to calculate the capacitance for any given frequency, assuming the inductance is known.

Step 4: Calculate Capacitance for Lower Frequency Limit

Using \( L = 220 \mu H = 220 \times 10^{-6} \text{ H} \) and \( f_1 = 800000 \text{ Hz} \), substitute into the formula: \( C_1 = \frac{1}{(2\pi \times 800000)^2 \times 220 \times 10^{-6}} \). Calculate this using a calculator.

Step 5: Calculate Capacitance for Upper Frequency Limit

Using \( f_2 = 1200000 \text{ Hz} \), substitute into the formula: \( C_2 = \frac{1}{(2\pi \times 1200000)^2 \times 220 \times 10^{-6}} \). Compute this using a calculator.

Step 6: Solve for C1 and C2

After calculating, find \( C_1 \approx 198 \text{ pF} \) and \( C_2 \approx 87.8 \text{ pF} \). These values represent the range of the variable capacitor.

Step 7: Compare with Choices and Select Answer

Compare the answers from Step 6 to the given multiple choices: (a) 87.8 to 198 pF matches the computed values. Thus, the correct range for the capacitor is 87.8 to 198 pF.

Key Concepts

Resonance Frequency

The resonance frequency in an LC circuit is an essential concept that determines how the circuit responds to different frequencies. It is the frequency at which the circuit can naturally oscillate due to the balance of energy between the inductor (storing energy in its magnetic field) and the capacitor (storing energy in its electric field). The formula for resonance frequency in an LC circuit is given by:

\[f = \frac{1}{2\pi \sqrt{LC}}\]where :

• \( f \) is the resonance frequency,
• \( L \) is the inductance measured in henries,
• \( C \) is the capacitance measured in farads.

This frequency is where the circuit can tune into a particular radio signal. In radio receivers, tuning to the correct frequency ensures the best reception. For our exercise, the radio can adjust between 800 kHz and 1200 kHz, using the resonance frequency to reach these desired signals efficiently. Understanding this allows you to appreciate why radios can focus on one station amidst multiple signals being broadcasted.

Inductance

Inductance in an LC circuit plays a crucial role in determining the resonance frequency. It corresponds to the circuit's ability to store energy in the form of a magnetic field when electrical current flows through it. The unit of inductance is henry (H), and it's denoted by \( L \) in equations.

In our problem, the effective inductance given is a constant value of 220 \mu\text{H} (microhenries). Microhenries are a subunit of henries, where :

• \( 1 \mu\text{H} = 1 \times 10^{-6} \text{ H} \).

The value of the inductance affects the overall tuning of the circuit, influencing both the lower and upper bounds of the frequency range the circuit can accommodate. With a fixed inductance of 220 \mu\text{H}, you see how it drives the necessary adjustments in capacitance to modify the resonant frequency for different broadcasting signals. Thus, understanding inductance is key to manipulating LC circuits for precise tuning in applications like radio, telecommunications, and electronic oscillators.

Capacitance Range

The capacitance range in an LC circuit determines its versatility and ability to tune into different frequencies within a specified range. In the exercise, we are tasked to find this range for tuning a radio that can operate between 800 kHz and 1200 kHz.Knowing the formula:

\[C = \frac{1}{(2\pi f)^2 L}\]allows us to calculate the required capacitance \( C \) as frequency changes. Here, the frequency \( f \) affects how much energy the capacitor can store in its electric field, influencing the resonance abilities.

• At the lower frequency limit, 800,000 Hz (800 kHz), the capacitance required is calculated to be approximately 198 pF (picofarads).
• At the upper frequency limit, 1,200,000 Hz (1200 kHz), the capacitance reduces to around 87.8 pF, showing how the required capacitance decreases as frequency increases.

The calculated range, from 87.8 pF to 198 pF, signifies the variability the capacitor must offer, ensuring the circuit can handle the whole frequency span. By understanding this concept, you appreciate how electronic devices can be fine-tuned for various inputs by merely adjusting capacitance within its designed range.